Question

The sum of 6 consecutive odd numbers is 20. What is the fourth number in this sequence?

Answer 1

There is no such sequence of `6` consecutive odd numbers.

Explanation:

Denote the fourth number by `n`.

Then the six numbers are:

`n-6, n-4, n-2, color(blue)(n), n+2, n+4`

and we have:

`20 = (n-6)+(n-4)+(n-2)+n+(n+2)+(n+4)`

`color(white)(20) = (n-6)+5n`

`color(white)(20) = 6n-6`

Add `6` to both ends to get:

`26 = 6n`

Divide both sides by `6` and transpose to find:

`n = 26/6 = 13/3`

Hmmm. That is not an integer, let alone an odd integer.

So there is no suitable sequence of `6` consecutive odd integers.

`color(white)()`
What are the possible sums of a sequence of `6` consecutive odd numbers?

Let the average of the numbers be the even number `2k` where `k` is an integer.

Then the six consectuvie odd numbers are:

`2k-5, 2k-3, 2k-1, 2k+1, 2k+3, 2k+5`

Their sum is:

`(2k-5)+(2k-3)+(2k-1)+(2k+1)+(2k+3)+(2k+5) = 12k`

So any multiple of `12` is a possible sum.

Perhaps the sum in the question should have been `120` rather than `20`. Then the fourth number would be `21`.